1. **Problem Statement:** We have a maintenance project with an initial cost of $100,000 in Year 1, increasing by $15,000 each year. The MARR (minimum attractive rate of return) is 8% per year. We want to find the present worth (PW) of the cash flows for: a) 5 years b) Infinite recurring cycles of 5 years each 2. **Formulas and Concepts:** - The cash flow is an arithmetic gradient series: initial amount $A_1=100000$, gradient $G=15000$. - Present worth of an arithmetic gradient series for $n$ years at interest rate $i$ is: $$PW = A_1(P/F,i,n) + G(P/A,i,n) - Gn$$ where: - $(P/F,i,n) = \frac{1}{(1+i)^n}$ is the present worth factor for a single future amount - $(P/A,i,n) = \frac{1-(1+i)^{-n}}{i}$ is the present worth factor for an annuity - For infinite repeating cycles of length $n$, the present worth is: $$PW_{infinite} = \frac{PW_{one\ cycle}}{(P/F,i,n)} - 1$$ 3. **Step-by-step for 5 years:** - Calculate present worth factors for $i=0.08$, $n=5$: - $(P/F,0.08,5) = \frac{1}{(1.08)^5} = 0.68058$ - $(P/A,0.08,5) = \frac{1-(1.08)^{-5}}{0.08} = 3.9927$ - Calculate PW: $$PW = 100000 \times 0.68058 + 15000 \times 3.9927 - 15000 \times 5 = 68058 + 59890.5 - 75000 = 52948.5$$ 4. **Step-by-step for infinite recurring 5-year cycles:** - The cash flow repeats every 5 years, so the infinite PW is: $$PW_{infinite} = \frac{PW_{one\ cycle}}{1 - (P/F,i,n)} = \frac{52948.5}{1 - 0.68058} = \frac{52948.5}{0.31942} = 165813.5$$ 5. **Summary:** - The present worth of the 5-year cash flow is approximately $52948.5$. - The present worth of the infinite recurring 5-year cycles is approximately $165813.5$. This approach uses the arithmetic gradient present worth formula and the geometric series formula for infinite repeating cash flows.